Rectangular telescoping of tensor details

Fix any one-dimensional input $g$ and coordinate $j$, and put $c_r:=Q_r(g)$. By [F3], ${\sum }_{r=1}^{m_j}(c_r-c_{r-1})=c_{m_j}-c_0$; for $m_j=0$ this is the empty-sum identity. Adding $Q_0(g)$ and using [F1] yields ${\sum }_{r=0}^{m_j}{\mathrm{\Delta }}_r(g)=Q_{m_j}(g)$. Since $g$ was arbitrary, this is equality of functionals.

For each coordinate $j$, represent every ${\mathrm{\Delta }}_r^{(j)}$, $0\le r\le m_j$, on the common finite node set $X_{m_j}^{(j)}$, inserting zero weights at absent nodes; this is possible by nestedness. Write its weight at $x$ as $a_{j,r}(x)$, so ${\sum }_{r=0}^{m_j}{\mathrm{\Delta }}_r^{(j)}$ has weight ${\sum }_{r=0}^{m_j}{a}_{j,r}(x)$ by scalar additivity. By [F1], the tensor of these coordinate sums has weight ${\prod }_{j=1}^d({\sum }_{r=0}^{m_j}{a}_{j,r}(x_j))$ at each node tuple. Repeated scalar scaling and additivity [F3], applied to the recursively formed product [F4], expand this weight as ${\sum }_{\mathrm{\ell }\in R_m}{\prod }_{j=1}^d{a}_{j,{\mathrm{\ell }}_j}(x_j)$. Multiplying by any input value $f(x_1,\dots ,x_d)$ and summing over the finite node tuples gives ${\sum }_{\mathrm{\ell }\in R_m}{\mathrm{\Delta }}_{\mathrm{\ell }}(f)$, since scalar additivity permits interchanging the finite sums. This is $Q_{R_m}(f)$ by [F2]. Step 1.1 identifies each coordinate sum with $Q_{m_j}^{(j)}$, proving the claimed equality of functionals for every $f$.